`4HCI + A to MnCl_2 + Cl_2 uarr + 2H_2O`, In this reaction the change in oxidation state of metal present in A is 

Options

A. `2` to `-2`
B. `+4` to `+6`
C. `-1` to `+3`
D. `+4` to `+2`

Answer

To determine the change in oxidation state of the metal present in compound A in the reaction \( 4HCl + A \rightarrow MnCl_2 + Cl_2 + 2H_2O \), we will follow these steps:

### Step 1: Identify the oxidation states of chlorine in the reactants and products.
- In \( HCl \), the oxidation state of Cl is -1.
- In \( Cl_2 \), the oxidation state of Cl is 0.

### Step 2: Determine the change in oxidation state for chlorine.
- The oxidation state of Cl changes from -1 in \( HCl \) to 0 in \( Cl_2 \). This indicates that \( HCl \) is being oxidized.

### Step 3: Identify the metal in compound A.
- The product \( MnCl_2 \) indicates that the metal in compound A is manganese (Mn).

### Step 4: Determine the oxidation state of manganese in \( MnCl_2 \).
- In \( MnCl_2 \), the oxidation state of Cl is -1. Therefore, for \( MnCl_2 \):
\[ X + 2(-1) = 0 \implies X - 2 = 0 \implies X = +2 \]
- Thus, the oxidation state of Mn in \( MnCl_2 \) is +2.

### Step 5: Determine the oxidation state of manganese in compound A.
- We need to deduce the oxidation state of Mn in compound A. Given that \( HCl \) is oxidized and Mn is reduced, we can assume that compound A contains manganese in a higher oxidation state.
- A reasonable assumption for compound A is \( MnO_2 \), where the oxidation state of Mn can be calculated as follows:
\[ X + 2(-2) = 0 \implies X - 4 = 0 \implies X = +4 \]
- Therefore, the oxidation state of Mn in \( MnO_2 \) is +4.

### Step 6: Calculate the change in oxidation state of manganese.
- The change in oxidation state of Mn from \( MnO_2 \) (+4) to \( MnCl_2 \) (+2) is:
\[ +4 \rightarrow +2 \implies \text{Change} = +4 - (+2) = +2 \]

### Conclusion:
The change in oxidation state of the metal present in A (Mn) is **2**.

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